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{\bf Question}

Using your matrix calculations from question 3 above, write down
the general solution to the following system of second order
differential equations:

$$\frac{d^2x}{dt^2}=2x + 3y$$

$$\frac{d^2y}{dt^2} = 2x + y.$$

{\bf Answer} \vspace{.2in}

In matrix form $\left( \begin{array}{c} \frac{d^2x}{dt^2} \\
\frac{d^2y}{dt^2} \end{array} \right) = \left( \begin{array}{cc} 2
& 3 \\ 2 & 1 \end{array} \right)\left(
\begin{array}{c} x \\ y
\end{array} \right)$

Eigenvalue $\lambda_1 = -1$ with eigenvector ${\bf x_1} = \left(
\begin{array}{c} 1 \\ -1 \end{array} \right)$ contributes a
solution $\left( \begin{array}{c} x \\ y
\end{array} \right) = \left( \begin{array}{c} 1 \\ -1
\end{array} \right) (A \cos t + B \sin t)$

Eigenvalue $\lambda_2 = 4$ with eigenvector ${\bf x_2} = \left(
\begin{array}{c} 3 \\ 2 \end{array} \right)$ contributes a
solution $\left( \begin{array}{c} x \\ y
\end{array} \right) = \left( \begin{array}{c} 3 \\ 2
\end{array} \right) (C e^{2t} + D e^{-2})$

The general solution to the equations is formed by adding these
two solutions:
\begin{eqnarray*}
\left( \begin{array}{c} x \\ y \end{array} \right) & = & \left(
\begin{array}{c} 1 \\ -1 \end{array} \right)
(A \cos t + B \sin t) + \left( \begin{array}{c} 3 \\ 2
\end{array} \right) (C e^{2t} + D e^{-2})\\ & = & \left(
\begin{array}{c} A \cos t + B \sin t + 3Ce^{2t} + 3De^{-2t} \\
-A \cos t - B \sin t + 2Ce^{2t} + 2De^{-2t} \end{array} \right)
\end{eqnarray*}
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